c++ - Are C macros implicitly cast?

Question

I've searched SO, but haven't found an answer to this specific questions. Forgive me if it's already been answered.

If you have the following:

#define MACRO  40

You don't assign it to a variable you use it in a loop:

for(int i = 0; i < MACRO; i++) {...

The perprocessor then creates:

for(int i = 0; i < 40; i++) {...

Would the compiler then implicitly cast it to an int since the comparison is with type int i? I've looked at this question Type of #define variables, and quite a few answers down Edgar Bonet implies that there is an order in which the compiler chooses how to treat the macro?

This question, How does C++ implicitly cast arguments to a comparator such as <?, was also suggested, but only describes how implicit casting works with a comparison with two types. Since a macro doesn't really have a type I'm not sure if this applies.

Answer 1

In C and C++, macros are quite literally in-place replacement. The preprocessor will encounter these #defines and replace them as it finds them. That's how you can nest macros inside of macros and it only takes 1 pass to preprocess.